Lemma 37.29.2. Let $f : X \to Y$, $s : Y \to X$ be as in Situation 37.29.1. If $g : Y' \to Y$ is any morphism, consider the base change diagram
so that we obtain $(X')^0 \subset X'$. Then $(X')^0 = (g')^{-1}(X^0)$.
Proof. Let $y' \in Y'$ with image $y \in Y$. We may think of $X^0_ y$ as a closed subscheme of $X_ y$, see for example Morphisms, Definition 29.26.3. As $s(y) \in X^0_ y$ we conclude from Varieties, Lemma 33.7.14 that $X_ y^0$ is a geometrically connected scheme over $\kappa (y)$. Hence $X_ y^0 \times _ y y' \to X'_{y'}$ is a connected closed subscheme which contains $s'(y')$. Thus $X_ y^0 \times _ y y' \subset (X'_{y'})^0$. The other inclusion $X_ y^0 \times _ y y' \supset (X'_{y'})^0$ is clear as the image of $(X'_{y'})^0$ in $X_ y$ is a connected subset of $X_ y$ which contains $s(y)$. $\square$
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